Problem: There are $9$ students in a class: $5$ boys and $4$ girls. If the teacher picks a group of $3$ at random, what is the probability that everyone in the group is a girl?
Answer: One way to solve this problem is to figure out how many different groups there are of only girls, then divide this by the total number of groups you could choose. Since every group is chosen with equal probability, this will be the probability that a group of all girls is chosen. We know two ways to count the number of groups we can choose: we use permutations if order matters, and combinations if it doesn't. Does the order the students are picked matter in this case? It doesn't matter if we pick Julia then Beatrice or Beatrice then Julia, so order must not matter. So, the number of ways to pick a group of $3$ students out of $9$ is $ \dfrac{9!}{(9-3)!3!} = \binom{9}{3}$ . [ Show me why Remember that the $9! \;$ and $(9-3)! \;$ terms come from when we fill up the group, making $9$ choices for the first slot, then $8$ choices for the second, and so on. In this way, we end up making $9\cdot8\cdot7 = \dfrac{9!}{(9-3)!} \;$ . The $3! \;$ term comes from the number of times we've counted a group as different because we chose the students in a different order. There are $3! \;$ ways to order a group of $3$ , so for every group, we've overcounted exactly that many times. We can use the same logic to count the number of groups that only have girls. Specifically, the number of ways to pick a group of $3$ students out of $4$ is $ \dfrac{4!}{(4-3)!3!} = \binom{4}{3}$ So, the probability that the teacher picks a group of all girls is the number of groups with only girls divided by the number of total groups the teacher could pick. This is $ \frac{\frac{4!}{(4-3)!\cancel{3!}}} {\frac{9!}{(9-3)!\cancel{3!}}} = \frac{\frac{4!}{1!}}{\frac{9!}{6!}}$ We can re-arrange the terms to make simplification easier $\left(\dfrac{4!}{1!}\right) \left(\dfrac{6!}{9!}\right) = \left(\dfrac{4!}{9!}\right) \left(\dfrac{6!}{1!}\right)$ Simplifying, we get $ \left(\dfrac{\cancel{4!}}{9\cdot8\cdot7\cdot6\cdot5 \cdot \cancel{4!}}\right) \left(\dfrac{6\cdot5\cdot4\cdot3\cdot2 \cdot \cancel{1!}}{\cancel{1!}}\right) = \left(\dfrac{1}{15120}\right) \left(720\right) = \dfrac{1}{21}$